题目原文：

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

题意解析：

给定两个链表表示两个非负数。数字逆序存储，每一个节点包括一个单一的数字。计算两个链表表示的数的和。并以相同格式的链表的形式返还结果。

解法就是直接操作链表相加就能够了。。

代码例如以下：

C++

class Solution {

public:

	ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

		ListNode * ans = NULL, *last = NULL;

		int up = 0;

		while (NULL != l1 && NULL != l2) {

			int tmp = l1->val + l2->val + up;

			up = tmp / 10;

			if (NULL == last) {

				ans = new ListNode(tmp % 10);

				last = ans;

			} else

			last = pushBack(last, tmp % 10);

			l1 = l1->next;

			l2 = l2->next;

		}

		while (NULL != l1) {

			int tmp = l1->val + up;

			last = pushBack(last, tmp % 10);

			up = tmp / 10;

			l1 = l1->next;

		}

		while (NULL != l2) {

			int tmp = l2->val + up;

			last = pushBack(last, tmp % 10);

			up = tmp / 10;

			l2 = l2->next;

		}

		if (0 != up) {

			ListNode * l = new ListNode(up);

			last->next = l;

		}

		return ans;

	}

	ListNode * pushBack(ListNode * last, int val) {

		ListNode * l = new ListNode(val);

		last->next = l;

		return l;

	}

};

Python

# class ListNode:

#     def __init__(self, x):

#         self.val = x

#         self.next = None

class Solution:

    # @return a ListNode

    def addTwoNumbers(self, l1, l2):

        carry = 0; head = ListNode(0); curr = head;

        while l1 and l2:

            Sum = l1.val + l2.val + carry

            carry = Sum / 10

            curr.next = ListNode(Sum % 10)

            l1 = l1.next; l2 = l2.next; curr = curr.next

        while l1:

            Sum = l1.val + carry

            carry = Sum / 10

            curr.next = ListNode(Sum % 10)

            l1 = l1.next; curr = curr.next

        while l2:

            Sum = l2.val + carry

            carry = Sum / 10

            curr.next = ListNode(Sum % 10)

            l2 = l2.next; curr = curr.next

        if carry > 0:

            curr.next = ListNode(carry)

        return head.next

水一枚。。。。